# Derivative of Inverse Function

• sydneyfranke
In summary, the student attempted to solve a homework equation that involved f(x) = -1/2. However, they got lost and needed help determining what g(-1/2) was. They plugged in f(x) into the inverse equation and got three solutions. They then found the derivative of the inverse at y=-1/2 and solved for y.
sydneyfranke

## Homework Statement

Find g' (-1/2), where g(x) is the inverse of f(x) = x3 / (x2 + 1)

## Homework Equations

g'(x) = 1/f' [g(x)]

## The Attempt at a Solution

Everything else I tried got really messy, but I feel like I'm going out on a limb by:

1) y=x3 / (x2+1)

2) y=x3x-2 / 1

3) y=x

4) g '(0.5)= 1/ (0.53 /(0.52+1) = 10

I don't even know if I'm going about it the right way. Any help would be awesome. Thanks.

You say in 1) that
y = x^3 / (x^2 + 1)
and in 3) that
y = x.

If you put in x = 1 you see that in 1) you get 1/2 and in 3) you get 1. Something must hence be wrong.

Very true. I realize something is wrong. That was just my attempt.

But . . . I could still use some help with this problem, obviously.

sydneyfranke said:
1) y=x3 / (x2+1)

2) y=x3x-2 / 1

Line 2 is incorrect, it should be

y=x3x-2 / (1+x-2),

which actually just makes things more complicated. Just differentiate the expression on line 1 directly.

It would help to figure out what g(-1/2) was equal to

Then you could just plug stuff into this relationship: g'(x) = 1/f' [g(x)] -> g'(-1/2) = 1/f' [g(-1/2)]

Since we know that g is the inverse of f, we should try to solve when f(a) = -1/2. Then we would know a = g(-1/2).

So -1/2 = x^3 / (x^2 + 1) -> -1/2*(x^2 + 1) = x^3 -> 0 = x^3 + 1/2x^2 + ½ which clearly has a solution at -1. So we know g(-1/2) = -1 which means: g'(-1/2) = 1/f'(-1) and you can do the rest from there id bet.

So I differentiate to get

(x2+1(3x2))-(2x(x3))/((x2+1)2)

And then I dared to simplify it down to

(x4+3x2)/((x2+1)2)

But what I don't understand from all of this is that if I do that, am I plugging this in for the g(x) for the 1/ f '[g(x)]? That would be like plugging f '[f '(x)] right? I'm sorry I'm just trying to understand how to solve these types of problems.

As the guy above said, if f(c) = -1/2 then g(-1/2) = c. To find out what g(-1/2) is we need to find c, that is to solve the equation f(x) = -1/2.

When this is done you can just plug everything into the formula.

sydneyfranke said:
So I differentiate to get

(x2+1(3x2))-(2x(x3))/((x2+1)2)

And then I dared to simplify it down to

(x4+3x2)/((x2+1)2)

But what I don't understand from all of this is that if I do that, am I plugging this in for the g(x) for the 1/ f '[g(x)]? That would be like plugging f '[f '(x)] right? I'm sorry I'm just trying to understand how to solve these types of problems.

Surely if I have a function y=f(x), then the inverse is just x=g(y) and it's derivative is $\frac{dx}{dy}$ which is a function of y. So that for:

$$f(x)=\frac{x^3}{x^2+1}$$

we have $x=f^{-1}(y)$ and so the derivative of the inverse at $y_0$ can be written as:

$$\frac{df^{-1}}{dy}\biggr|_{y_0}=\frac{dx}{dy}\biggr|_{y_0}=\frac{1}{\frac{dy}{dx}}\biggr|_{x(y_0)}$$

So now let $y_0=-1/2[/tex]. Then solve: $$y_0=-1/2=\frac{x^3}{x^2+1}$$ which has three solutions. The inverse and it's derivative are multi-valued right since it's going to be a cube root correct? Therefore, just find the three roots [itex] (x_1,x_2,x_3)$ and for example, then one derivative of the inverse at y=-1/2 is:

$$\frac{df^{-1}}{dy}\biggr|_{y=-1/2}=\frac{dx}{dy}\biggr|_{y=-1/2}=\frac{1}{\frac{dy}{dx}}\biggr|_{x(-1/2)=x_1}$$

Try and follow the following Mathematica code which explicitly computes the inverse, then finds it's (multivalued) derivative at y=-1/2:

Code:
f[x_] := x^3/(x^2 + 1);
mysols = x /. Solve[f[x] == y, x]
N[D[mysols, y] /. y -> -2^(-1)]

Last edited:

## 1. What is the derivative of an inverse function?

The derivative of an inverse function is the reciprocal of the derivative of the original function.

## 2. What is the notation used for the derivative of an inverse function?

The notation used for the derivative of an inverse function is f'(x) or dy/dx.

## 3. How do you find the derivative of an inverse function?

To find the derivative of an inverse function, you can use the inverse function theorem which states that the derivative of an inverse function is equal to 1 divided by the derivative of the original function evaluated at the inverse function's input.

## 4. Can the derivative of an inverse function be negative?

Yes, the derivative of an inverse function can be negative. This means that the original function is decreasing at that point.

## 5. Why is the derivative of an inverse function important?

The derivative of an inverse function is important because it allows us to find the slope of a curve at any point, and also helps us to solve problems involving rates of change and optimization.

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